如何永远运行2个异步功能？

问题描述

我正在尝试永远运行2个async函数。有人能帮我吗？下面提供了我的代码和错误消息。

代码：

import websockets
import asyncio
import json
import time

async def time(loop):
    while True:
        millis = await int(round(time.time() * 1000))
        print(millis)
        await asyncio.sleep(0.001)

async def stream(loop):
    async with websockets.connect('wss://www.bitmex.com/realtime?subscribe=trade:XBTUSD') 
                    as websocket:
        while True:
            try:
                data = await websocket.recv()
                data = json.loads(data)
                print(data['data'][-1]['price'])
            except KeyError:
                pass
            except TypeError:
                pass

loop = asyncio.get_event_loop()

async def main():
    await asyncio.gather(loop.run_until_complete(stream(loop)),loop.run_until_complete(time(loop)))

if __name__ == "__main__":
    asyncio.run(main())

错误：

Exception has occurred: RuntimeError
Cannot run the event loop while another loop is running

解决方法

好，您的代码段错误很少：

1. 您不能将第一个函数命名为time，因为它会与time内置函数产生冲突
2. 为什么不使用loop作为参数呢？这没用。
3. 如果无法等待某个功能，则无法await，即int是一个内置的同步方法。

进行正确的更正将是这样的：

import websockets
import asyncio
import json
import time

async def another_name():
    while True:
        millis = int(round(time.time() * 1000))
        print(millis)
        await asyncio.sleep(0.001)

async def stream():
    async with websockets.connect('wss://www.bitmex.com/realtime?subscribe=trade:XBTUSD') 
                    as websocket:
        while True:
            try:
                data = await websocket.recv()
                data = json.loads(data)
                print(data['data'][-1]['price'])
            except KeyError:
                pass
            except TypeError:
                pass
            await asyncio.sleep(0.001) #Added

if __name__ == "__main__":
    loop = asyncio.get_event_loop()
    coros = []
    coros.append(another_name())
    coros.append(stream())
    loop.run_until_complete(asyncio.gather(*coros))

await asyncio.sleep(0.001)函数中的行stream()是强制性的，否则它将永远不会让another_name()函数运行。